Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			final int r = sc.nextInt();
			final int c = sc.nextInt();

			int[][] g = new int[r][c];
			for (int i = 0; i < r; i++)
				for (int j = 0; j < c; j++)
					g[i][j] = sc.nextInt();

			StringBuilder sb = new StringBuilder();

			int moves = 0;
			if (c > r)
				moves += reduceRows(g, sb) + reduceCols(g, sb);
			else
				moves += reduceCols(g, sb) + reduceRows(g, sb);

			if (complete(g)) {
				System.out.println(moves);

				System.out.println(sb.toString());
			} else {
				System.out.println(-1);
			}
		}
	}

	static int reduceRows(int[][] g, StringBuilder sb) {

		int modif = 0;

		// for every row
		for (int r = 0; r < g.length; r++) {
			// find the minimum and reduce all by it
			int min = Integer.MAX_VALUE;
			for (int c = 0; c < g[r].length; c++)
				if (g[r][c] < min)
					min = g[r][c];

			for (int c = 0; c < g[r].length; c++)
				g[r][c] -= min;

			modif += min;

			for (int i = 0; i < min; i++) {
				sb.append("row ");
				sb.append(r + 1);
				sb.append("\n");
			}
		}

		return modif;
	}

	static int reduceCols(int[][] g, StringBuilder sb) {

		int modif = 0;

		// for every column
		for (int c = 0; c < g[0].length; c++) {
			// find the minimum and reduce all by it
			int min = Integer.MAX_VALUE;
			for (int r = 0; r < g.length; r++)
				if (g[r][c] < min)
					min = g[r][c];

			for (int r = 0; r < g.length; r++)
				g[r][c] -= min;

			modif += min;

			for (int i = 0; i < min; i++) {
				sb.append("col ");
				sb.append(c + 1);
				sb.append("\n");
			}
		}

		return modif;
	}

	static boolean complete(int[][] g) {
		for (int r = 0; r < g.length; r++)
			for (int c = 0; c < g[r].length; c++)
				if (g[r][c] > 0)
					return false;
		return true;
	}

	static void printG(int[][] g) {
		for (int i = 0; i < g.length; i++)
			System.out.println(Arrays.toString(g[i]));
	}

}